HDU1061 Rightmost Digit(快速幂取余)

最新推荐文章于 2019-04-28 22:27:38 发布

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#快速幂

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本文介绍了一种解决HDU1061题目中快速计算N^N最右位数的方法，通过快速幂取余算法，有效解决了大整数运算的问题，避免了直接计算带来的溢出。

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题目: HDU1061

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 68789 Accepted Submission(s): 25687

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

2 3 4

Sample Output

7 6

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

方法:

此题使用快速幂取余来解决。

注意有两个取模运算公式：

1. (a * b) % c = [(a % c) * (b % c)] % c，即积的取余等于取余的积的取余;

2. $a^b\,\%\,c = (a\,\%\,c)^b\,\%\,c$

代码:

#include <bits/stdc++.h>
using namespace std;

poww(int a, int b, int c)
{
	int ans = 1;
	a = a % c;
	while (b > 0)
	{
		if (b % 2 == 1)
			ans = (ans * a) % c;
		b = b / 2;
		a = (a * a) % c;
	}
	
	return ans;
}

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int n;
		cin >> n;
		cout << poww(n, n, 10) << endl;
	}
	
	return 0;
}

关于快速幂取余的讲解，点击此处到达。